Question

If $3x-\frac{1}{3x}=15$, find the value.

$81x^4+\frac{1}{81x^4}$

• A. $51527$
• B. $51529$
• C. $225$
• D. $227$

Answer 1

The correct option is A $51527$

$81x^4+\frac{1}{81x^4}=?$

$3x-\frac{1}{3x}=15\left(Given\right)$

Squaring both sides, we have

${(3x-\frac{1}{3x})}^2={\left(15\right)}^2$

$\Rightarrow {\left(3x\right)}^2+{\left(\frac{1}{3x}\right)}^2-2\times 3x\times \frac{1}{3x}=225$

$\Rightarrow 9x^2+\frac{1}{9x^2}-2=225$

$\Rightarrow 9x^2+\frac{1}{9x^2}=225+2$

$\Rightarrow 9x^2+\frac{1}{9x^2}=227$

Again squaring both sides, we have

${(9{}^2-\frac{1}{9x^2})}^2={\left(227\right)}^2$

$\Rightarrow {\left(9x^2\right)}^2+{\left(\frac{1}{9x^2}\right)}^2+2\times 9x^2\times \frac{1}{9x^2}=51529$

$\Rightarrow 81x^4+\frac{1}{81x^4}+2=51529$

$\Rightarrow 81x^4+\frac{1}{81x^4}=51529-2$

$\Rightarrow 81x^4+\frac{1}{81x^4}=51527$

Hence, the correct answer is $51527$.